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This is problem number 33 of the stewart calculus 8th edition section 2 .4.
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Verify that another possible choice of delta for showing that the limit as x approaches 3 of the function x squared is equal to 9.
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In example 4 is delta equals minimum of 2 and epsilon over 8.
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So, recall that the epsilon delta definition of a limit is that if the, the, the difference between x and a, the absolute value of the difference is less than delta, then the absolute value of the difference between the function and the limit is epsilon, less than epsilon, provided that epsilon and delta are both greater than zero.
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So we begin and just use this definition directly and use our delta values.
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Let's see if x, the absolute value of x minus a, a is 3, is less, than we're going to use this first delta assuming that this is the minimum, so less than 2, then if this is true, we can rewrite it as negative 2, less than x minus 3, less than, and what we're going to do is we're going to convert this to x plus 3 instead in the middle.
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And in order to do this, it means adding 6 to each term, 6 minus 2, or 6 plus negative 2 is 4.
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6 plus 2 is 8.
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And in order to modify this and change it back to an absolute value, we see that x plus 3 is less than 8.
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X plus 3 is also greater than 4.
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X plus 3 is also greater than any value less than 4...